php outputting code rather than executing - php

Question

I am pretty new to php, but have to link a database to html through php as part of uni coursework, so I attempted to implement a register and log in feature to the site we created. As far as I am aware this code should work however whenever I try to run it through the html page it posts the code itself rather than running. This is the case for both the login and registration. What am I doing wrong?
//registration
<?php
define('DB_HOST', 'xxx');
define('DB_NAME', 'xxx');
define('DB_USER','xxx');
define('DB_PASSWORD','xxx');
$con = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect to MySQL: " . mysql_error());
$db = mysql_select_db(DB_NAME,$con) or die("Failed to connect to MySQL Database: " . mysql_error());
function NewUser() {
$Name = $_POST['Name'];
$Username = $_POST['Username'];
$Email = $_POST['Email'];
$Password = $_POST['Password'];
$query = "INSERT INTO userDetails (Username,Email,Name,Password) VALUES ('$Username','$Email','$Name','$Password')";
$data = mysql_query($query) or die(mysql_error());
if($data) {
echo "YOUR REGISTRATION IS COMPLETED...";
}
}
function SignUp() {
if(!empty($_POST['Username'])) {
$query = mysql_query("SELECT * FROM userDetails WHERE Username = '$_POST[Username]' AND Password = '$_POST[Password]'") or die(mysql_error());
if(!$row = mysql_fetch_array($query) or die(mysql_error())) {
newuser();
} else {
echo "YOU ARE ALREADY A REGISTERED USER...";
}
}
}
if(isset($_POST['submit'])) {
SignUp();
}